The New Sudoku Players' Forum

[TKiel]

Shurd12,

Using the xy-wing noted by Hud as an example, r9c1 can't be 2 because that would make r7c3=7 and r1c1=5, which would leave r8c1 with no candidates.

With as many (2,5), (2,7), (2,9), (5,7) (7,9) pairs floating around in this grid as there are, it seems there aught to be a way to make eliminations with xy-wings just about anywhere one chooses. Or just go cross-eyed from looking, which is what happened to me. So I found even another way to solve it.

From the grid as first posted, after making eliminations in r8c19 from the (2,9) naked pair in row 8, in r9c1 from the (2,7:7,5:5,2) xy-wing use by Hud, in r4c7 from the locked candidate 5's in r8 box 6, in r4c8 from the (7,9) naked pair now in row 4, in r4c3 from the locked candidate 2's in column 3 box 7, the candidate grid looks like this:

Code: Select all

 
 *--------------------------------------------------*
 | 25   9    3    | 27A  4    6    | 1    8    57   |
 | 1    7    4    | 8    5    3    | 2    9    6    |
 | 6    25   8    | 1    9    27a  | 57A  4    3    |
 |----------------+----------------+----------------|
 | 3    25   1    | 79a  6    4    | 79A  25   8    |
 | 279  6    79   | 5    8    1    | 479  3    247  |
 | 8    4    579  | 3    2    79   | 6    57   1    |
 |----------------+----------------+----------------|
 | 4    3    27   | 6    1    5    | 8    27   9    |
 | 57   8    6    | 29   3    29   | 457  1    457  |
 | 59   1    259  | 4    7    8    | 3    6    25   |
 *--------------------------------------------------*


Colouring on 7's shows two (A) in column 7, so (a) must be true.

Tracy

[frans mulder]

5 9 3 2 4 6 1 8 7
1 7 4 8 5 3 2 9 6
6 2 8 1 9 7 5 4 3
3 5 1 7 6 4 9 2 8
2 6 9 5 8 1 7 3 4
8 4 7 3 2 9 6 5 1
4 3 2 6 1 5 8 7 9
7 8 6 9 3 2 4 1 5
9 1 5 4 7 8 3 6 2

[frans mulder]

in case you are interested I have a simple excel help that solves all puzzles within two minutes
Frans

[ravel]

Frans,
most people here have a program, that solves them in milliseconds. See also this

[Hud]

ib2064,

Code: Select all

Don't think that works. Both branches must be able to eliminate the target candidate. If r8c1 is 5 then r1c1 becomes a 2 but it cannot eliminate r9c3.

You're right. I guess I was more tired than I thought lol.

[shurd12]

An xy-wing is a 3-cell chain where each cell has two candidates in it. The chain looks something like ab-bc-ca, where the bc cell sees both the ab and the ca cells. It is easy to prove in a chain like this, that one end or the other must equal 'a'. Therefore, any cell which sees both endpoints of the chain cannot be an 'a'.

How does r1c1 "see" r7c3? I still don't get it.

[shurd12]

Shurd12,

Using the xy-wing noted by Hud as an example, r9c1 can't be 2 because that would make r7c3=7 and r1c1=5, which would leave r8c1 with no candidates.

This I understand. Thanks.

[shurd12]

How about this: If r1c4 is a 7, r4c4 is a 9, r4c7 is a 7, r3c7 is a 5 and r1c8 is a 7. This can't be so r1c4 is a 2.

[Sped]

Starting with what shurd12 posted:

Code: Select all

 *-----------*
 |.93|.46|18.|
 |174|853|296|
 |6.8|19.|.43|
 |---+---+---|
 |3.1|.64|..8|
 |.6.|581|.3.|
 |84.|32.|6.1|
 |---+---+---|
 |43.|615|8.9|
 |.86|.3.|.1.|
 |.1.|478|36.|
 *-----------*

 
 *-----------------------------------------------------------*
 | 25    9     3     | 27    4     6     | 1     8     57    |
 | 1     7     4     | 8     5     3     | 2     9     6     |
 | 6     25    8     | 1     9     27    | 57    4     3     |
 |-------------------+-------------------+-------------------|
 | 3     25    1     | 79    6     4     | 579   257   8     |
 | 279   6     279   | 5     8     1     | 479   3     247   |
 | 8     4     579   | 3     2     79    | 6     57    1     |
 |-------------------+-------------------+-------------------|
 | 4     3     27    | 6     1     5     | 8     27    9     |
 | 2579  8     6     | 29    3     29    | 457   1     2457  |
 | 259   1     259   | 4     7     8     | 3     6     25    |
 *-----------------------------------------------------------*


After excluding on the naked pair in row 8, and eliminating the 5 in r4c7 on locked candidates, we get:

Code: Select all

 
 *--------------------------------------------------*
 | 25   9    3    | 27   4    6    | 1    8    57   |
 | 1    7    4    | 8    5    3    | 2    9    6    |
 | 6    25   8    | 1    9    27   | 57   4    3    |
 |----------------+----------------+----------------|
 | 3    25   1    | 79   6    4    | 79   257  8    |
 | 279  6    279  | 5    8    1    | 479  3    247  |
 | 8    4    579  | 3    2    79   | 6    57   1    |
 |----------------+----------------+----------------|
 | 4    3    27   | 6    1    5    | 8    27   9    |
 | 57   8    6    | 29   3    29   | 457  1    457  |
 | 259  1    259  | 4    7    8    | 3    6    25   |
 *--------------------------------------------------*


The XY Chain:

2-(r1c4)-7-(r4c4)-9-(r4c7)-7-(r3c7)-5-(r3c2)-2

eliminates the 2s at r1c1.and r3c6.

It's all singles from there.

[Myth Jellies]

An xy-wing is a 3-cell chain where each cell has two candidates in it. The chain looks something like ab-bc-ca, where the bc cell sees both the ab and the ca cells. It is easy to prove in a chain like this, that one end or the other must equal 'a'. Therefore, any cell which sees both endpoints of the chain cannot be an 'a'.

How does r1c1 "see" r7c3? I still don't get it.

Code: Select all

 *--------------------------------------------------*
 |*25*  9   -3    | 27   4    6    | 1    8    57   |
 | 1    7   -4    | 8    5    3    | 2    9    6    |
 | 6    25  -8    | 1    9    27   | 57   4    3    |
 |----------------+----------------+----------------|
 | 3    25   1    | 79   6    4    | 79   257  8    |
 | 279  6    279  | 5    8    1    | 479  3    247  |
 | 8    4    579  | 3    2    79   | 6    57   1    |
 |----------------+----------------+----------------|
 |-4    3   *27*  | 6    1    5    | 8    27   9    |
 |*57   8    6    | 29   3    29   | 457  1    457  |
 |-259  1    259  | 4    7    8    | 3    6    25   |
 *--------------------------------------------------*


r1c1 is the first cell, which sees the second cell r8c1, which sees the last cell r7c3. Note that a = 2, b = 5, and c = 7, and you have the whole picture. You can eliminate all 2's in cells that see both endpoints, in this case the cells I have marked with minus signs.

[shurd12]

Got it. Thanks.

[Cec]

in case you are interested I have a simple excel help that solves all puzzles within two minutes
Frans

Hi frans mulder,
I'm sure you think you are helping by posting the solution but the forum is for "helping to solve" the puzzle. Simply posting the solution does nothing to teach anyone how to solve it. There are ample solver programs available which can provide an immediate solution to a puzzle but few of us, if any, can get excited doing that. If you look back through the forum's posts you'll see how solving techniques are posted and you are of course welcome to do that but please refrain from posting future solutions.
Cec

[ravel]

Frans, i forgot to mention that your excel post should go to the Solver Programs forum, where you also can find other Excel programs (try a "Search"), if you are interested.

[Cec]

The XY Chain:

2-(r1c4)-7-(r4c4)-9-(r4c7)-7-(r3c7)-5-(r3c2)-2
eliminates the 2s at r1c1.and r3c6.
It's all singles from there.

Hi Sped,
Your above notation refers to five cells and six candidates (2,7,9,7,5 and 2). Could you please explain your interpretation of this notation.
Cec

[ravel]

The XY Chain:
2-(r1c4)-7-(r4c4)-9-(r4c7)-7-(r3c7)-5-(r3c2)-2
eliminates the 2s at r1c1.and r3c6.

Strange notation, but i think, it is clear, what is meant:
Either r1c4=2
or r1c4=7, r4c4=9, r4c7=7, r3c7=5, r3c2=2